Term Rewriting System R:
[N, X, Y, XS]
fib(N) -> sel(N, fib1(s(0), s(0)))
fib1(X, Y) -> cons(X, fib1(Y, add(X, Y)))
add(0, X) -> X
add(s(X), Y) -> s(add(X, Y))
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, XS)

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

FIB(N) -> SEL(N, fib1(s(0), s(0)))
FIB(N) -> FIB1(s(0), s(0))
FIB1(X, Y) -> FIB1(Y, add(X, Y))
FIB1(X, Y) -> ADD(X, Y)
ADD(s(X), Y) -> ADD(X, Y)
SEL(s(N), cons(X, XS)) -> SEL(N, XS)

Furthermore, R contains three SCCs. 


   R
     ↳DPs
       →DP Problem 1
         ↳Polynomial Ordering
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Remaining


Dependency Pair:

SEL(s(N), cons(X, XS)) -> SEL(N, XS)


Rules:


fib(N) -> sel(N, fib1(s(0), s(0)))
fib1(X, Y) -> cons(X, fib1(Y, add(X, Y)))
add(0, X) -> X
add(s(X), Y) -> s(add(X, Y))
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, XS)





The following dependency pair can be strictly oriented:

SEL(s(N), cons(X, XS)) -> SEL(N, XS)


There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(SEL(x1, x2))=  x1  
  POL(cons(x1, x2))=  0  
  POL(s(x1))=  1 + x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 4
             ↳Dependency Graph
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Remaining


Dependency Pair:


Rules:


fib(N) -> sel(N, fib1(s(0), s(0)))
fib1(X, Y) -> cons(X, fib1(Y, add(X, Y)))
add(0, X) -> X
add(s(X), Y) -> s(add(X, Y))
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, XS)





Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polynomial Ordering
       →DP Problem 3
         ↳Remaining


Dependency Pair:

ADD(s(X), Y) -> ADD(X, Y)


Rules:


fib(N) -> sel(N, fib1(s(0), s(0)))
fib1(X, Y) -> cons(X, fib1(Y, add(X, Y)))
add(0, X) -> X
add(s(X), Y) -> s(add(X, Y))
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, XS)





The following dependency pair can be strictly oriented:

ADD(s(X), Y) -> ADD(X, Y)


There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(s(x1))=  1 + x1  
  POL(ADD(x1, x2))=  x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
           →DP Problem 5
             ↳Dependency Graph
       →DP Problem 3
         ↳Remaining


Dependency Pair:


Rules:


fib(N) -> sel(N, fib1(s(0), s(0)))
fib1(X, Y) -> cons(X, fib1(Y, add(X, Y)))
add(0, X) -> X
add(s(X), Y) -> s(add(X, Y))
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, XS)





Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Remaining Obligation(s)




The following remains to be proven:
Dependency Pair:

FIB1(X, Y) -> FIB1(Y, add(X, Y))


Rules:


fib(N) -> sel(N, fib1(s(0), s(0)))
fib1(X, Y) -> cons(X, fib1(Y, add(X, Y)))
add(0, X) -> X
add(s(X), Y) -> s(add(X, Y))
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, XS)




Termination of R could not be shown.
Duration:
0:00 minutes